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%I #20 Mar 30 2024 09:16:56
%S 1,2,3,30,5,9,7,50,20,42,11,36,13,6510,27,54,17,70620,19,25,46,66,23,
%T 168630,124,98,58,78,29
%N Index of first occurrence of n in A109883, or -1 if n does not occur in A109883.
%C Sequence continues: a(29)=??, a(30)=??, 31, 70, 112, 100, 57, 200, 37, 484, 55, 102, 41, 49, 43, a(44)=??, 94, 114, 47, 225, 1264, 252, 104, 294, 53, 780, 87, 71940, 118, 138, 59, 4290, 61, 1470, 85, 306, 134, a(66)=??, 67, 6300, 142, 288, 71, 324, 73, a(74)=??, 712, 174, 158, 2940, 79, a(80)=??, 166, 186, 83, 1344210, 405, 242, 115, 1590, 89, a(90)=??, 141, 196, 406, 540, 119, 2310, 97, 390, 202, 222, ..., . - _Jason Earls_ and _Robert G. Wilson v_, Jul 12 2005
%C Smallest number N with perfect deficiency n, that is, the first number such that A109883(N)=n. - _Walter Kehowski_, Sep 13 2005
%C a(29) > 10^9 if it exists. - _Michel Marcus_, Mar 30 2024
%e a(7) = 50: divisors of 50 are 1,2,5,10,25,50; 50 - (1 + 2 + 5 + 10 + 25) = 7 and 50 is the smallest such number.
%p with(numtheory); pdef := proc(n) local k, d, divd; d:=n; divd:=sort([op(divisors(n))]); for k in divd while d>=k do d:=d-k; od; end: PDL:=[]; for z from 1 to 1 do for pd from 0 to 60 do for n from 1 to 200000 do missed:=true; if pdef(n)=pd then PDL:=[op(PDL),n]; missed:=false; break fi od; if missed then PDL:=[op(PDL),-1] fi od od; PDL; # _Walter Kehowski_, Sep 13 2005
%t subtract = If[ #1 < #2, Throw[ #1], #1 - #2]&; f[n_] := Catch @ Fold[subtract, n, Divisors @ n] (* Bobby R. Treat, (DrBob(AT)bigfoot.com), Jul 14 2005 *)
%t t = Table[0, {60}]; Do[ a = f[n]; If[a < 60 && t[[a + 1]] == 0, t[[a + 1]] = n], {n, 10^8}] (* _Robert G. Wilson v_, Jul 14 2005 *)
%Y Cf. A064510, A109883, A109884.
%K nonn,more
%O 0,2
%A _Amarnath Murthy_, Jul 11 2005
%E Corrected and extended by _Jason Earls_, Jul 12 2005
%E More terms from _Walter Kehowski_, Sep 13 2005
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